# Every prime ideal has finite height in a Noetherian ring

In Corollary 11.12 of Atiyah-Macdonald, it says that in a Noetherian ring every prime ideal has finite height. It seems to come directly from Proposition 11.10, which says if $$A$$ is a Noetherian local ring, $$\dim A \leq d(A)$$, where $$d(A)$$ is the degree of the characteristic polynomial of $$A$$, hence finite. It seems like from that proposition you can only conclude that any prime ideal in a Noetherian local ring has finite height. I don't see how you can reach Corollary 11.12.

• If you believe in Corollary 11.11, then recall that the height of a prime ideal $\mathfrak p$ equals $\dim A_{\mathfrak p}$ (which is a noetherian local ring). – user26857 Jan 5 '20 at 22:51
• How would a chain of ideals establishing infinite height not violate the ascending chain condition? (asks a non-algebraist) – Eric Towers Jan 5 '20 at 23:08
• @EricTowers: You don't need an infinite ascending chain to get infinite height, just arbitrarily long finite chains. – Eric Wofsey Jan 5 '20 at 23:11
• @EricTowers Also, the height of a prime ideal $\mathfrak{p}$ is the supremum of lengths of chains of prime ideals contained in $\mathfrak{p}$, so the height of a prime ideal has something to do with the descending chain condition, not the ascending chain condition. – Emily Williams Jan 6 '20 at 20:59

However, the height of any prime ideal $$\mathfrak{p}$$ in a Noetherian ring $$R$$ can still never be infinite. To see this, as the comments mention, one can combine Proposition 11.10 in Atiyah-Macdonald with the fact that $$\operatorname{ht}\mathfrak{p}=\dim R_{\mathfrak{p}}$$, which follows from their Corollary 3.13, to conclude that $$\operatorname{ht}\mathfrak{p}$$ is finite.